1

Variantim 79 וריאנטים

December 2019

מקוריות: פתרונים והערות

Originals: Solutions & Comments

IRT judges: #2: Valery Kopyl (2019) #3: Jiří Jelínek (2018-9) #n: Gerhard E.Schoen (2018-20) Studies:

Steffen Slumstrup Nielsen (2019) H#2-3: Ricardo Vieira; H#>3:Aleksandr Semenenko (2019); S#: Genady

Kozyura (2019-20) Fairies: Hans Gruber (2019)

Editors: :עורכים Orthodox: Ofer Comay

Studies: Gady Costeff (Please send originals in pgn format)

Fairies: Michael Grushko

ofercomay@gmail.com

costeff@gmail.com

bargrushko@bezeqint.net

עופר קומאיבעיות רגילות:

גדי קוסטףסיומים: (pgnקוריות בפורמט )נא לשלוח מ

מיכאל גרושקובעיות אגדתיות:

3303

Zoltan Labai

Slovakia

#2*vvv 9+9

1…e5 2.Qxd7#

1…dxc6 2.Qe6#

1.Be5+ ? 1…Kxe5 ! 1…fxe5 2.Qd3# (A)

1.Bxf6 (B) ? 2.Qd3# (A) 1…exf6 !

1.Qf3 ? 2.Sf7# 1…cxd4 !

1…e5 2.Qd5#

1…dxc6 2.Qxc6#

1.Qd3 ! (A) [2.Bxf6# (B)]

1…e5 2.Be3#

1…cxd4 2.Qxd4#

1…Bxc3 2.Bxc3#

Zagorujko [1…e5] changed mates [1… dxc6] Urania [A] reversal 2 [AB-BA]

(author)

I think there are possibilities here to avoid the crude refutations of both thematic

tries (PE)

3304

Valery Shanshin

Russia

#2v 9+11

1.Bxe7(A)? [2.Qf4(B)#]

1…Sxh6(a) 2.Rb5(C)#

1…Kxe4(b) 2.Re3#

but 1…Sxe4!

1.Rb5(C)! [2.Qxe6#]

1…Sxh6(a) 2.Bxe7(A)#

1…Kxe4(b) 2.Qe3# [2.Qf4(B)? Kd3!]

1…Sf6 2.Qf4(B)#

Reversal-I (Salazar), anti-Dombrovskis and pseudo-Le Grand (author)

The anti-Dombrovskis feature (2.Qf4 is not a mate after 1…Kxe4) is an amusing

element in contemporary two-movers (PE)

3305

Hubert Gockel

Germany

#2vvvv 11+4

1.Qb3? [2.c4#] 1.- Qxd4 x 2.cxd4# A 1.- Qxb6!

1.Rb3? [2.c4#] 1.- Bxd4 y 2.cxd4# A 1.- Qb8,Qb6 2.e5# 1.- Qa5!

1.Qe5? [2.Sg1#] 1.- Qxd4 x 2.Sxd4# B 1.- Qxg5!

1.Qf7? [2.Sg1#] 1.- Qxd4 x 2.Sxd4# B 1.- Qxg5 2.Bxg5# 1.- Qf6!

1.Rf6! [2.Sg1#] 1.- Bxd4 y 2.Sxd4# B 1.- Qxg5 2.e5#

Reciprocal changes of two mates spread over 4 phases. One extra phase(1.Qf7?)

doubles the xB occurrence, improves usage of wBf4 and obtains a changed mate

after 1.- Qxg5 (author)

A very nice execution of the “two black lines” mechanism of the 4-phase

reciprocal changes theme (PE)

'd'H'd'd dP0p0'd' '0Pip0'd d'0'd'd' 'dPG'd'd d')'d'dQ 'd'd')Kd g'd'd'd'

rd'd'd'd 0'd'0'd' Kd'dpd'! d'G'ipdp 'd'dPdn$ dRd'd'h' BdNdpdNd g'd'd'd'

'd'1'd'd g'd'd'd' '$'d'dpd d'dQd')' 'd')PG'd d')'dkdP 'd'dNd'd dBd'dKd'

2

3306

Givi Mosiashvili

Pavel Murashev

Russia

#2vvv 8+9

1...Kxc5 (a) 2.Be3# 1...Bc~ (c) 2.Qxd5# (A)

1.Qh3? [2.Qe3# (X)] 1...Kxc5!

1...Ke4 2.Rxc4# (B) - self-pin (2...dxc4?)

1.Re6? [2.Be3# (C)] 1...Bh6!

1...Kxc5 (a) 2.Qxd5# (A) - self-pin (2...Bxd5?)

1.Rxb4? [2.Qxd5# (A)] - pin, 1...Rxa8!

1...Kxc5 (a) 2.Rcxc4# (B) - self-pin (2...dxc4?)

1...Rxc5 (b) 2.Qxd3# (Y) - pin

1.Bxd5! [2.Rxc4# (B)]

1...Kxc5 (a) 2.Be3# (C) - self-pin (2...d4?)

1...Rxc5 (b), Bc~ (c) 2.Qe4#

1...Bxd5! 2.Qxd3# (Y)

1.Qe6? [2.Qe3 (X), Be3# (C)] 1...Kxc5! (a)

Barnes, Dombrovskis paradox (aC) 1.Rd6? .2.Rxc4# (B) - pin, 1...Rxc5!

1...Kxc5 (a) 2.Qxd5# (A) - self-pin (1...Bc~ 2.R,Qxd5#) Le Grand (AB-BA)

B7/6b1/1R3p2/r1=Pp1=Q2/1pbk1=B2/1p1p4/ 1N5K/2R5 Ukrainian theme (CA-

AB-BC) + self-pinning (3 thematic variants! - new) �œ Changed (abc) and

transferred (ABY) mates, Shedey theme (B) �œ Rukhlis theme, complete (*aC,

cA .aA .bY .bZ, zY) �œ Black correction in solution. (authors)

The main idea seems to me the Ukrainian cycle:

1.Rxb4? [2.Qxd5 A#] 1...Kxc5 a 2.Rcxc4 B# but 1...Rxa8!

1.Bxd5! [2.Rxc4 B#] 1...Kxc5 a 2.Be3 C#

1.Re6? [2.Be3 C#] 1...Kxc5 a 2.Qxd5 A# but 1...Bh6!

It is based mainly on pins, unpins and self-pins, and is excellent in itself. The

Dombrovskis paradox elements are actually an inherent part of the Ukrainian

cycle but are combined with a “Barnes try” (meaning two threats separated to

single threats in other phases). Amazingly, there is also a Rukhlis theme hiding

among the phases:

1.Rxb4? 1...Rxc5 b 2.Qxd3 Y#

1.Bxd5! 1...Rxc5 b 2.Qe4# 1...Bxd5 2.Qxd3 Y#

1...Kxc5 a 2.Be3 C#

1.Re6? 1...Kxc5 a 2.Qxd5 A#

Set play 1...Bc~ c 2.Qxd5 A#

3307

Bogusz Piliczewski

Poland

#2vvvv 10+5

1.Sf~? [2.Qxf6#] 1....Kxf4! (2.Qh2?)

1.Sd4? 1....Kxf4 2.Qh2# 1....Sxf4! (2.Bc3?)

1.Sfd6? 1....Sxf4 2.Bc3# 1....Sg7! (2.Bc7?)

1.Sg7? 1....Se~ 2.Bc7# 1....Sxf4 2.Bc3# 1....f5 2.Sxe6#

1....Kxf4 2.Qh2# 1....Sg~ 2.R(x)e4# 1....Sxf3! (2.Qf6?)

1.Sfe7! [2.Qxf6#] 1....Sg7 2.Bc7# 1....Sxf4 2.Bc3# 1....Kxf4 2.Qh2#

1....Sh7/e4 2.R(x)e4#

White correction - White knight wheel - Line closings (author)

The indicated try 1.Sf~? is actually the combination of 1.Sh6,1.Sh4,1.Sg3? all

interfering with the WQ mate on h2, so since these moves are harmful to white

there’s no real random move here. Even without this, the problem is interesting,

especially the way a zugzwang play is introduced after 1.Sg7? (PE)

3308

Yosi Retter

Ganei Tikva

#2* 8+5

1...fxe3 a 2.Bd5 A#

1...fxg3 b 2.Bh5 B#

1.Rh2 ! zugzwang.

1...fxe3 a 2.Bh5 B#

1...fxg3 b 2.Bd5 A#

I suggests to move WRh4 to h2 this will create the tries:

1.Re2? fxg3! 1.Rg2? fxe3! (David Stern)

This reciprocal changes mechanism is well known, but it is Ntly done here in

mutate form (PE).

Bd'd'd'd d'd'd'g' '$'d'0'd 4')pdQd' '0bi'G'd dpdpd'd' 'H'd'd'I d'$'d'd'

'dNd'd'! d'd'd'd' 'd'dn0Pd G'd'iNh' 'dBd'$'d d'd'0Pd' 'd'dPd'd I'd'd'd'

'd'd'$'d d'd'hBh' 'd'd'd'd d'0'd'd' 'dKd'0'$ d'd'HkH' 'd')'d'd d'd'G'd'

3

3309

Gerard Doukhan

France

#2*vvv 10+10

Set Play: 1....Bxc5 [a], Kc3 2.Qb2 [A] #

1.B6 ~ ? [2.Qb2 [A] #] 1....Sb3 2.Sxb3# but 1....Bc7!

1.Sc ~ ? [2.Qb2 [A] #]

1....e3 [b] 2.Qd3 [B] # ]

but 1....Sb3!

1.Sxe4? [2.Qd3 [B] #] Threat Correction

1....Be2 [c] 2.Qb2 [A] #

1....dxe4 2.Qc4# [C]

but 1....Sf2!

1.Bxd5! [2.Qc4# [C] Threat Correction

1....exd5 2.Qb2 [A] #

1....Kxd5 2.Qd7# 1....Be2 2.Sxe6#

Double threat correction, Cycle threat / mat AB, BC, CA Maybe the first example

presenting the two themes in association (author)

Nice threat correction play with a key granting an additional flight (PE)

3310

Zoltan Labai

Miroslav Svitek

Slovakia/Czech Rep.

#3 13+8

1.Ka5 ! [2.Sc4+ (A) Kxe4 3.Qb1#]

1…Qxd6 (a) 2.Bd4+ (B) Kxe4 3.Qb1#

1…Qxe4 (b) 2.Bd4+ (B) Qxd4 3.Qxd4#

1…Sc3 2.Sxc3 [3.Sc4# (A)] 2…Qxd6 (a) 3.Bd4# (B)

2…Qxe4 (b) 3.Re6# (C)

1…Bc7 2.Qxc7 [3.Bd4# (B)] 2…Qxd6 (a) 3.Re6# (C)

2…Qxe4 (b) 3.Qe7# (D)

Replies to 1st move defenses by the BQ remain the same also in two additional

variations when the BQ defenses are in the 2nd move and the replies are mates

(PE)

3311

Aleksandr Sygurov

Russia

In memoriam

Arieh Grinblat

#3 8+12

1.Qe3! zz;

1...de2 2.Rxf3! (zz) Qg2/Sxf3 3.Qxe2#

1...fe2 2.Qxd3! (zz) Qg2/Sf3 3.Qxe2#

1...Sxe2 2.Rxf3! Qg1 3.Qxf2#

1...Qxe2 2.Rgxg1+ f1~ 3.Rxf1#, 2…fg 3.Rxg1#

1...Qg2 2.ef3+ Kf1 3.Bxd3#

Half-pin & pin-mates, pinning of e2 defences on same square (e2) (author)

Nice half-pin play (PE)

3312

Yitzhak Nevo

Ein Harod

#3 10+6

1.Sg4 ! [2.Se5 + Rxe5 3.Rd4# 2...Rxe5 3.Rd4# 2...Bxe5 3.Bd5#] 1...Rxb5

2.Ba6 [3.Bxb5#] 2...Re5 3.Rd4# 2...Bd4 3.Rxd4#

1...Bc3 2.Rc7 + Rc5 3.Bd5# 2...Rc6 3.Se3#

1...Rc5 2.Bxc5 [3.Bd5#] 2...Re5 3.Rd4# 2...Rc6 3.Se3# 2...Rd6 3.Se3# 2...Bd4

3.Rxd4#

1...Re4 2.Bxe4 [3.Se3#] 2...Rxg4 3.Bd5# 2...Bd4 3.Rxd4#

It seems that BPb2 is not needed. Also, the distant WSh2 suggests the solution. I

suggest moving WSh2 to g4 and WRd2 to d5. Now the position can be moved

one row down (WBf1) the solution is 1.Rd1 and the rest is the same (David Stern)

The two anti-critical defenses are of interest (PE)

'dRd'd'd 0'd')'d' 'gBdpI'd 0QHpH'd' Pd'ipdb$ d'd'd'd' 'd'd'd'd h'G'd'dn

'g'd'd'4 d'd'dBd' P!pH'dq$ d'GPi')b 'I'dPdpd d'd'd')' 'd'dNdPd d'dnd'd'

'd'd'd'd 0'd'd'd' Bd'dQd'd d'd'd'd' 'd')'d'd d'dpdp$p '0'0P0') dKgriqhR

'd'd'd'd dBd'd'gR '0'drd') dPd'd'4' K)kd'd'd d'd'd'd' N0'$'d'H d'd'd'G'

4

3313

Leonid Makaronetz

Viktor Volchek

Haifa/Belarus

#3 12+7

1.Sfxe5! [2.Bf3+Kd4 3.Re4#]

1…Raxe5 2.Sc5+Kd4 3.Rd8#

1…Rhxe5 2.Rf1 Kd4 3.Rf4#

1…Rh1 2.Sf3+Re5 3.Re5#

1…Kd4 2.Sc5 Rc5 3.Sf3#

Flight-giving key, self-pins & pin-mates and two quiet variations. An entertaining

mix (PE)

3314

L.Lyubashevsky

L.Makaronetz

Rishon LeZion/Haifa

#3v 11+7

1.cxb3? [2.Qb1+ d3 3.Qxd3#]

1...Be3 2.Qa5 [3.Qe5#] 2...Bf4 3.Qd5#

1...Bh2(d6) 2.Qd2 [3.Qxd4#] 2...Bg1 3.Sg3# 2...Be5 3.Sxg5# 2...d3 3.Qe3# but

1...d3!

1.Qb4! [2.Qc5 ~ 3.Qxd4(d5) # 2...Be3 3.Qe5#]

1...Bd6 2.Qd2 [3.Qxd4#] 2...Bc5 3.Sg3# 2...Be5 3.Sxg5#

1...Be5 2.Sxg5 + Kf4 3.Qd2 #

1...Sf6 2.Qe7 + Be5 3.Qxe5 #

Some changed continuations and good quiet variations (PE)

3315

Gerard Doukhan

France

#3* 13+11

Set Play: 1...Bxd5 [a] 2.Bd3# [A] 1...Sxd4 [b] 2.Sg5# [B]

1.Rb5? [2.Bd3# [A]]

1....Se5 2.Bxe5 [3.Bd3, Sg5#] but 1...Bxd5! [a]

1.Bc5? [2.Sg5# [B]] but 1...Sxd4! [b]

1.Rf6! [2.Rxe6+ Se5 3.Rxe5#]

1...e5 2.Rb5 ! [3.Bd3# [A]] 2...Bxd5 [a] 3.Bxd5#

1...Sf8 2.Bc5 ! [3.Sg5# [B]] 2...Sxd4 [b] 3.Rxf4# 2...Sh7 3.Rxe6# 2...gxf6

3.Sxf6#

Dombrovskis theme (author)

The Dombrovskis between the set play and tries is changed in the solution into

something else. Very interesting (PE)

3316

Yosi Retter

Ganei Tikva

#3* 11+8

1...Rxf5 a 2.Sd3 + A Bxd3 3.Qa1#

1...Bxf5 b 2.Sf3 + B Rxf3 3.Qa1#

1.Qxh7! [2.Qxf7]

1...Rxf5 a 2.Sf3+ B 2.Rxf3 3.Sd7#

1...Bxf5 b 2Sd3+ A Bxd3 3.Sd7#

Very clear reciprocal changes. In the set play the two black pieces clear the 1st

row while in the solution the piece that captured on f5 is moved back to open a

guard on e4 (PE)

'd'dRd'd d'd'd'd' 'dPd'G'0 4'd'0'dr ')PdkdPd d'dNdN0' pd')Bd'd I'd'$'d'

'd'd'dnI d'0Bd')' 'dPd'd'd d'd'dN0' 'dP0kg'd dpd'dNdP 'dPdPd'd d'd'!'d'

b4'd'$'d d'd')'0' ')'Gpdnd d'dPdndN 'dB)k0P0 dR0'0Nd' Kd'dPd'd d'd'd'd'

'dKd'd'd d'd'dpdp 'd'dpH'd d')'iP)' 'dPd'dPd d'd')'0' 'd'dpdPd dbd'HrdQ

5

3317 A.Sygurov, Russia

In memoriam

Arieh Grinblat

#4 9+12

1.Rd5 ! zugzwang.

1...Scxe2 2.Rxd6 zz Kc1 3.Rxd3 zz Se~ 4.Qd2# (2...Sgxe2? 2...Bxe2?

2...fxe2?)

1...Bxe2 2.c4 zz Kd1 3.Bc3 [4.Qe1#] 3...d2 4.Rxd2# (2...Scxe2? 2...fxe2?

2...fxe2?)

1...Sgxe2 2.Rd4 zz Kxc3 3.Qe1 + d2 4.Qxd2# 2...d5 3.Bf8 zz Kxc3 4.Bb4#

(2...Scxe2? 2...Bxe2? 2...fxe2?)

1...fxe2 2.Sg3 [3.Sf1,Se4#] 2...fxg3 3.Bh6 + Kxc3 4.Qd4# (2...Scxe2?

2...Bxe2?)

Task: incarceration in four variations (author)

Each of the captures on e2, with self-pins, causes all other black pieces to be

immovable. The quiet replies all lead to interesting play (PE)

3318

Michael Pasman

Meitar

#7 10+10

White is preparing the maneuver Re8 followed by moving the Bishop from c2 to

h3/g4 and at the end mate Re6. But meanwhile black promotes Queen, So there

are some obstacles...

1.d4! (1.Nc7? Kf5 2.d4+ (2.dxe3 Kg4 ) 2...Kg4) (1.Nxb4? cxb4 2.dxe3 a2 )

1...bxa6 - First blockage 2.d5! Second blockage and also opening a4-d7 diagonal

(2.dxe3? Be1 3.d5 cxd5 4.exf4 d4 5.e4 Bxh4 6.Bb3 c4 ) 2...cxd5 3.Ba4!

(3.Nc3? Bxc3 4.Ba4 exd2) (3.dxe3? Be1! 4.exf4 d4 5.e4 Bxh4) 3...a2

(3...Bxd2=Important line 4.Bd7 Be1 5.Bg4 Bxh4 6.Re6+ Kg5 7.Rg6#! )

4.Nc3! Third blockage (4.Bd7? a1=Q 5.Bg4 Qe5 ) 4...Bxc3 (4...a1=Q 5.Nxd5+

Kf5 6.Bd7# ) 5.Bd7! a1=Q 6.Bg4! Fourth blockage. Now the line a is closed

by a6 pawn, the diagonal a2-g8 is closed by pawn d5, diagonal a1-e5 is closed by

Bishop c3, and g-line is closed by Bishop g4.The queen cannot prevent next

mate.[#] 6...Qg1 or any other move 7.Re6#

Crisp logic: the maneuver Bc2-a4-d7-g4 will allow Re6# but black will promote a

queen. White pre-arranges blocks for the black queen preventing Qa8 (pin of

WRe8), Qe5, Qa2+ & Qg1+. Excellent! (PE)

3319

Michael Pasman

Meitar

Draw 5+11

1.Qc2! Bf2 1...g3 2.Qxe2+ Bf2 3.Rc2 transposes 2.Qxe2 g3 3.Rc2 Ne6+

4.Kg4 f5+ 5.Kh4 Nf4 6.Qxf2+ gxf2 7.Rxf2+ Ng2+ 8.Rxg2+ Kxg2 9.b5 f4

10.b6! 10.bxa6? f3 11.a7 f2 12.a8=Q f1=Q 13.Qg8+ Kh1! 14.Qg3 Qc4+ 15.Kh3

Qe2 -+ 10...f3 11.b7 f2 12.b8=Q f1=Q 13.Qg3+ Kh1 14.Qf3+! 14.Qe3 Qf6+

15.Kg3 Qd6+ 16.Kf3 Qf8+ 17.Kg3 Qg7+ 18.Kh4! Qg2 19.Qxh6 Qf2+ 20.Kg4

d4 21.Qd6 Qg2+ 22.Kh4 Qe4+ 23.Kg5 d3 24.h6 Kg1 -+ 14...Qg2 15.Qd1+

Qg1 16.Qf3+ Qg2 17.Qd1+ Kh2 18.Qg1+ draw.

3320

Janos Mikitovics

Hungary

Draw 3+6

1.Qa6! 1.Qg1? Nxe4! ( 1...Kxe4? 2.Qb1+ Nd3 3.Qb7+ Nd5 4.Qh7+ Kd4

5.Qa7+ Kc4 6.Qa6+ Kd4 7.Qa7+ Nc5 8.Qg7+ Kd3 9.Qg5 Ke4 10.Qg6+ Kd4

11.Qg7+ Kc4 12.Qg4+ Kd3 13.Qg5 positional draw ) 2.Qh2+ Kf5 3.Qh3+ Ke5

4.Qxf3 a2 -+ 1...Nxe4 2.Qh6+ Ke5! 3.Qh8+ 3.Qg7+? Ke6 -+ ( 3...Kd6?

4.Qd4+ = ) 3...Ke6 ( 3...Kd5 4.Qd8+ = ( 4.Qh5+? Kc4 5.Qg4 Kd3 6.Qd7+ Ke2

7.Qxe7 f2 8.Qxe4 f1=Q -+ ) ) ( 3...Nf6 4.Qb8+ Ke6 5.Qb6+ Kf7 6.Kg3 )

4.Qa8! Nc3 5.Qxa3! 5.Qxf3? Ned5 -+ 5...f2! 6.Qa6+! 6.Qxc3? Nf5+ -+

6.Qb3+? Ned5 -+ 6...Ke5 7.Kh3!! 7.Kg3? Ne4+ ( 7...Ned5? 8.Kh2 Ne4 9.Qb5

main ) 8.Kh2 Nd2 -+ ( 8...Nf5? 9.Qb5+ Kf4 10.Kg2 Nh4+ 11.Kh3 Nf3 12.Qb8+

Ne5 13.Qf8+ = ) ( 8...Nd5? 9.Qb5 main ) 7...Ned5 8.Kh2!! 8.Kg2? Ne4 9.Qb5

e2!! 10.Qxe2 Nf4+ -+ 8...Ne4 8...e2 9.Qxe2+ Nxe2 10.Kg2 = 9.Qb5!! Nd2!

10.Qe8+ 10.Qb8+? Kd4! ( 10...Ke4? 11.Qe8+ main ) 11.Qa7+ Kd3 -+

10...Kd4 11.Qa4+ Kd3 12.Qb5+ Kc2 13.Qxd5 f1=Q 14.Qa2+ Kd3

15.Qa3+! Ke2 15...Kd4 16.Qa7+! Kd3 17.Qa3+ main 16.Qa6+ Ke1 17.Qa1+

Kf2 18.Qf6+ Nf3+ 19.Qxf3+! Kxf3 stalemate.

'd'd'd'd d'd'd'G' 'd'0'd'd d'd'd'$' 'd'd'0'd dp)pdpdp p$piP!') I'hbd'hN

'd'dRdKd dpd'd'd' Ndp0'i'0 d'0'd'dP 'g'd'0') 0'dP0'd' 'dB)Pd'd d'dNd'd'

'd'd'd'd d'dpd'dQ pdpd'0'0 d'dpd'hP ')Rd'Ipd d'd'd'd' 'd'dpd'i d'd'd'g'

'd'd'd'd d'd'h'd' 'd'd'd'd d'd'd'd' 'd'dPi'I 0'd'0pd' 'd'd'h'd d'd'dQd'

6

3321

Vladislav Tarasiuk

Ukraine

Win 6+6

See Kubbel Rigaer Tageblatt 1914 1.Nd3+! exd3 2.Bf2 exd4+ 3.Kd2!! Logical

try-1: 3.Kxd3? g1=Q 4.Bxg1 Kg3 5.Nc5 Kxh3! 6.Ne4 ( no moves 6.Nd3)

6...Kg2 7.Bf2 h3 = 3...g1=Q 4.Bxg1 Kg3 5.Nc5! Try: 5.Ne5? f2 6.Bxf2+

Kxf2 = 5...Kxh3! ( 5...f2 6.Ne4+ Kxh3 7.Bxf2 +- ) ( 5...Kg2 6.Bxd4 f2 7.Bxf2

Kxf2 8.Ne4+ Kg2 9.Ng5 Kg3 10.Kxd3 Kf4 Position arises from a study by L.

Kubbel 11.Ne4 Kf3 12.Kd4 Kf4 13.Kd5! Kf5 14.Nc3! Kf4 15.Ne2+ Kf3

16.Ng1+ Kg2 17.Ke4! Kxg1 18.Kf3 Kh2 19.Kg4 +- ) 6.Nxd3 Kg2 7.Bf2!!

Logical try-2: 7.Bxd4? f2! ( 7...h3? 8.Ne1+ Kh1 9.Nxf3 h2 10.Be5! +- ) 8.Bxf2

h3 9.Ke2 h2 10.Nf4+ Kh1 11.Kf1 stalemate 7...h3 8.Ke1 h2 9.Nf4+ Kh1

10.Kf1 d3 11.Nh5! 11.Nxd3? stalemate ) 11...d2 12.Ng3#

3322

Daniele Gatti

Italy

Win 12+9

1.Bxe7 e1=Q 2.d8=N+ Kxe7 3.Nc6+ Kd6 4.Rxf2 Kxc7 5.Re2 Qxe2

6.Bd7 Qxb5 7.Rc8+ Kxd7 8.Na7 Qxb7+ 9.Kxb7 Rxc8 10.Nxc8 d4

11.Nb6+ Ke6 12.Kc6 h5 13.f5+ gxf5 14.gxh5 wins.

3323

Branislav Djurasevic

Serbia Dedicated to Fadil

Abdurahmanovic

For his 80th birthday

Draw 3+4

1.Bh3! 1.h6? Kxf1 2.h7 Bd7! 3.h8=Q g3+ 4.Kxh1 Bc6# 1.Bg2? Bxh5! 2.Kxh1

Kf2! -+ 1...g3+! 1...gxh3 2.h6! Bg6 3.Kxh1 Kf2 4.Kh2 Bf5 5.h7 = 1...Bxh5

2.Bxg4! = 1...Nf2 2.h6! Bg6 3.Bg2 = 2.Kg1! 2.Kg2? Bxh5! -+ Thematic try:

2.Kxh1? Kf2! 3.Bg2 Bxh5! 4.Bd5 Bf3+ 5.Bxf3 Kxf3 -+ Position A. Mutual zz,

WTM. 2...Nf2! Main line B: 2...Bxh5 3.Bg2! Nf2 4.Bf3!! ( 4.Bc6? Bg6! 5.Kg2

Be4+ 6.Kg1 Nh3# ( 6...Bxc6? = ) ) 4...Bg6 ( 4...Bxf3 = stalemate ) 5.Kg2! Nh1

6.Kg1! Nf2 7.Kg2 Be4 8.Kxg3! 3.Bc8!! 3.Bf5? Bxh5! 4.Kg2 Nh1! 5.Be4 Ke2

6.Kg1 Nf2! -+ 3.Be6? Nd3! 4.Kg2 Nf4+ -+ 3.Bg2? Ke2 4.Bb7 Ke3! 5.Kg2

Kf4! -+ 3...Bxh5 3...Bc6 4.Bf5! ( 4.h6? Be4! -+ ) 4...Nd1 5.h6! Nc3 6.Bg4 =

4.Kg2! 4.Bb7? Bg6! 5.Bf3 ( 5.Kg2 Be4+ -+ ) 5...Be4! 6.Kg2 Bxf3+ -+

4...Nh1 5.Bb7! ( 5.Kg1? Bg6! ( 5...Bf3? 6.Bb7! Bxb7 = ) 6.Bb7 Nf2 -+ )

5...Ke2 6.Kg1! Nf2 7.Kg2! Nh1 8.Kg1 Bf3 9.Bxf3+! Kxf3 10.Kxh1 draw.

Position A. Mutual zz, BTM.

3324

E.Egorov

Pyotr Kiryakov

Kazakhstan/Russia

Win 4+5

1.Rb1 d4 2.cxd4 c3 3.Rf1! 3.Kh7? Kd5 4.Rf1 c2 5.Rc1 Kc4 6.Rxc2+ Kd3

7.Rc1 Kxe3 8.d5 Kd2 9.Rh1 e3 10.d6 e2 11.d7 e1=Q Black will have Qb1+

12.d8=Q+ Ke2 13.Qe7+ Kd2 14.Qd6+ Ke2 15.Qe5+ Kd2 16.Qd4+ Ke2!

17.Rh2 Qb1+! 18.Kh6 Qf5 = 3.Kg7? Kd5 4.Rf1 c2 5.Rc1 Kc4 6.Rxc2+ Kd3

7.Rc1 Kxe3 8.d5 Kd2 9.Rh1 e3 10.d6 e2 11.d7 e1=Q = 3.Rc1? Kd5 4.Rf1 c2 or

3.Kh8 c2 4.Rc1 ( 4.Rf1 Kd6 5.Rc1 Kd5! see 4.Rc1 ) 4...Kd5! 3...c2 4.Rc1!

4.Kh8? Kd6! 5.Rc1 Kd5! Mutual zugzwang with white to play! 4...Kd5 Black is

forced to play Kd5, if 4...Kd6 then at least 5.Kf7 with easy win - white king goes

to e4 pawn 5.Kh8!! Mutual ZZ is in White`s favor now! 5...Kc4 5...Ke6 the

white king goes to Pe4, for example 6.Kg7 Kd5 7.Kf6 Kc4 8.Rxc2+ Kd3 9.Rxf2

6.Rxc2+ Kd3 7.Rc1! 7.Rxf2 Kxe3 = 7...Kxe3 8.d5 Kd2 9.Rh1! Try 9.Ra1 e3

10.d6 e2 11.d7 e1=Q 12.d8=Q+ Ke2 13.Qe8+ ( 13.Ra2+ Kf1! ) 13...Kf1 14.Qe5

Qxa1! ) 9...e3 10.d6 e2 11.d7 e1=Q 11...f1=Q 12.d8=Q+ with a quick win, for

example 12...Kc2 13.Qc7+ Kd3 14.Qd6+ Ke3 15.Qe5+ Kf3 16.Qf5+ 12.d8=Q+

Ke2 13.Qe7+ Kd2 14.Qd6+ Ke2 15.Qe5+ Kd2 16.Qd4+ Kc2 16...Ke2

17.Rh3 17.Rh2! wins.

'd'd'd'd d'dNd'd' 'd'd'd'd d'd'0'd' 'd')pi'0 d'I'dpdP 'd'd'Hpd d'd'G'd'

K$BG'd'4 dP)P0kdp 'd'd'dpd dPdpd'd' 'd'd')Pd dpd'd'd' ')'dpgRd d'd'd'd'

'd'dbd'd d'd'd'd' 'd'd'd'd d'd'd'dP 'd'd'dpd d'd'd'd' 'd'd'd'I d'd'iBdn

'd'd'dKd d'd'd'd' 'd'dkd'd dRdpd'd' 'dpdpd'd d')')'d' 'd'd'0'd d'd'd'd'

7

3325

Yochanan Afek

Janos Mikitovics

Amsterdam/Hungary

Draw 4+5

1.d4+! 1.Rxc4+ $2 Kb5 1...Kd5 1...Kxd4 2.Rxc4+ Kd3 3.Rxb4 = 1...Kb5

2.bxc4+ Ka4 3.Kb2 b3 4.Kc3 b2 5.Kxb2 = 1...Kb6 2.bxc4 Be6 3.Kb3 = 2.bxc4+

2.Rxc4 $2 Ra7+ 3.Kb2 Rb7 4.Rc5+ Ke4 5.Re5+ Kf4 -+ 2...Kxd4 3.Kb3 Rb7

4.Rd1+ 4.c5 $2 Be6+ -+ 4...Ke3 4...Bd3 5.c5 $1 Rb5 6.c6 = 4...Ke4 5.Rd6 Bh7

6.Ra6 Kd4 7.Ra4 Bg8 8.Rxb4 = transposes to main line 5.Rd6 5.Rd5 Bd7 6.c5

Be6 pin -+ 5...Bh7 5...Be4 6.c5 = 6.Ra6 6.Rd8 Bg6 7.Rd5 Bf7 -+ 6.Rd5 Bg8

6...Bg8 7.Ra4 7.Ra8 $2 Be6 8.Ra4 Kd4 9.Rxb4 Bxc4+ 10.Ka3 loss of time

7...Kd4 8.Rxb4 Bxc4+ 9.Ka3 Ra7+ 10.Ra4 10.Kb2 Kd3 11.Rb8 Ra2+ -+

10...Rf7 11.Rb4 Rf3+ 12.Ka4 ( 12.Kb2 Rh3 13.Rb8 Rh2+ 14.Ka3 Ra2+

15.Kb4 Rb2+ -+ 12...Rf2 13.Ka5 Kc5 14.Rb5+ Bxb5 stalemate.

3326

Vladislav Nefyodov

Vitaly Medintsev

Russia

H#2 3.1.1.1 10+14

1.Bf2 (Bh2?) Rh1 2.Kxd4 Rh4#

1.Rg8 (Rg6?) Bh5 2.Kxd5 Bxf7#

1.b5 (b6?) Ra6 2.Kxc5 Rc6#

Three matching solutions in which black’s 1st move must avoid interfering with

the mating piece. From my point of view, the fact that in one solution (1.Bf2)

there is line opening while in the other two unguards, increases the problem’s

quality. White’s strategy is also interesting (PE)

3327

Emanuel Navon

Holon

H#2 2.1.1.1 10+10

1.Scxd5 (1.Sa4?Sb5?) Qxf7 (1...Qd8? 2.Se3?) 2.Sb6 Bb1#

1.Sfxd5 (1.Sg4?) Qd8 (1...Qxf7? 2.Sb6?) 2.Se3 Sf4#

The main question after each capture on d5 is how white should continue. After

1.Scxd5 1...Qd8? is not good as c3 is unguarded and 2…Sf4 is not a mate.

Similarly, after 1.Sfxd5 1...Qxf7? will not work as BSc3 guards b1 and 2…Bb1

is not a mate (PE)

3328

Emanuel Navon

Holon

H#2 b)Bf2h2 9+9

a) 1.Sc5 e6 2.Bd4 Sb6#

b) 1.Bd3 g5 2.Rd4 Sd2#

Composed originally for TT 227 Superproblem 2019. Theme: The second Black

move (B2) has two negative effects, for instance a check to white king and a

capture of a needed white unit. One of the negative effects needs to be overcome

(neutralized) in the first Black move (B1), and the other - in the first White move

(W1). At least two phases with given theme condition (author)

The move 2.Bd4 will open the h6-a6 row and directly guard b6. 1.Sc5 is an

anticipatory interference of Bd4 and 1…e6 and anticipatory interference of

BQh6. A similar strategy in b): 1.Bd3 is an anticipatory interference of Rd4 and

1…g5 and anticipatory interference of BQh6 on the 6-c1 diagonal. Very nice (PE)

'd'd'd'd d'd'drd' 'd'd'd'd d'i'dbd' '0pd'd'd dPd'd'd' Kd')'d'd d'$'d'd'

'dnd'd'd dpd'dp4' 'd'd'0'd $')NdK)' PdkH'd'd d'd'0B0' p0pdPd'd 4bdRd'g'

bd'd'dQd drd'dp0' 'dKd'hpd d')P)'d' 'dPHpd'd d'hkd'd' Bd'gN)'d d'd'd'd'

'd'd'd'd d'dnd'd' 'd'd'g'1 $'dN)'d' p)kdb4Pd )'d'd'd' 'dBd'4nd dNd'd'I'

8

3329

Emanuel Navon

Holon

H#2 b)Re2a2 7+10

1.Sg3 (~?) S8f7 (Kf6?) 2.Rxd4 Re5#

1.Sb3 (~?) Ra4 (Rc2,d2?) 2.Bxd6 c7#

Also composed for Superproblem TT 227 (2019)

In a) 1.Sg3 is an anticipatory interference while 1…S8f7 provide a guard on e5,

anticipating the unguard by 2.Rxd4. In b) 1.Sb3 is an anticipatory interference

while 1…Ra4 provide a guard on c4, anticipating the unguard by 2.Bxd6.

Harmonious and interesting! (PE)

3330

Shaul Shamir

Rishon LeZion

H#2 2.1.1.1 8+4

1.Qc5 Be5 (Rf3?) 2.Qa3 d3# (d4?)

1.Ra6 Rf3 (Be5?) 2.Ra3 d4# (d3?)

One of the versions of Shaul’s 9th Prize in the 2019 Sabra tourney. After black’s

decision to block a3 with either the rook or queen white must choose how to

guard c3 – a choice that determines if WPd2 will move one or two squares (PE).

3331

Viktor Yuzyuk

Ukraine

H#2 4.1.1.1 6+10

1.Kc5 Qxd2 2.Kc4 Qxc3#

1.Ke5 Qe3 2.Sd6 f4#

1.Ke7 Qxd4 2.Kf8 Qg7#

1.Kc7 Qxc3 + 2.Kb8 Qc8#

BK moves only ª 3 Cross (Q) Echo mates (rotated 90, 1, 4) Many ways (Q, 2) Star

(k) Echo Chameleon (author)

BK start & WQ cross – nicely done (PE)

3332

Michal Dragoun

Czech Rep.

H#2 6.1.1.1 9+11

1.a2 Qxa2 2.Kd3 Sf2#

1.b2 Qc2 + 2.Kf3 Sh2#

1.Ba5 Sfe3 2.Kd4 Qxc4#

1.Rb6 Sge3 2.Ke5 Qe8#

1.Kxf5 Qa7 2.Re4 Qf7#

1.Kd5 Qxb5 2.Re6 Qb7#

One of the developments of the modern helpmate was to take the HOTF concept

further with three (or more) pairs of solutions. Here all three pairs are nicely

connected and comprise an artistically pleasing problem (PE)

3333

Menachem Witztum

Tel Aviv

H#2 3.1.1.1 10+12

1.d1=R cxb3 2.Rd3 Sc2#

1.bxc5 Sg4 2.c4 Qa7#

1.Bf6 Qxg6 2.Bxe5 c3#

Cyclic change of functions (c2,e3,g7) (author)

Pawn c2 guards c4 to enable Sd3 to mate - Sd3 guards e5 to enable the WQ to

mate – WQ guards d3 to enable pawn c2 to mate. Nicely done (PE)

Bd'H'dbd d'd'd'I' pdPH'd'd d'hkdpd' 'g')n4'd d'd'd'd' 'd'dRd'1 drd'd'd'

'd'I'd'd d'd'd'd' 'd'drd'd d'd'd'dq 'dPd'G'd d')'d'd' Bi')'dRd g'd'dRd'

'd'g'd'd 0ndPd'd' 'd'ibdKd dpdpd'd' 'd'hPd'd d'4Qd'd' ')'1')'d d'd'd'd'

Kd'd'd'd d'd'd'0' 'd')'d'd dr0'dN)' QgpdkdN) 0p0'd')P 'd'd'd'd d'1'4'd'

'd'd'd'd d'd'd'!b '0')pdpd d')p)'gn ')'i'0'd dpd'HPh' 'IP0')'d d'd'd'd'

9

3334

Daniel Papack

Germany

H#2 2.1.1.1 4+9

1.Rxg5 Sxh2 2.Rf5 Qg4#

1.Bxg4 Bxh6 2.Bf5 Qg5#

For mating on g4 the WB must be removed so g6 will be guarded. For mating on

g5 the WS must be removed to allow the WQ access. Harmonious solutions with

reciprocal annihilations (PE)

3335

Krzysztof Drążkowski

Poland

H#2 2.1.1.1 4+10

1.Kc5 Sf3 2.Bd5 d4#

1.Ke5 d3 2.Rd5 Sf3#

Nice repeat of the WS move to “match” the non-repeating WP moves (PE)

3336

Zoltan Labai

Slovakia

H#2.5 3.1.1.. 6+10

1...Bc3 2.Ra4 + Ba5 3.Rb4 Bxb6#

1...Rd2 2.Re5 Rxd6 3.Rd5 Rc6#

1...Rc3 2.Kd4 g3 3.Ke5 Rxe3#

The 3rd solution, not showing a switchback and differing substantially from the

other two, seems to me not in place (PE)

3337

Vitaly Medintsev

Russia

H#2.5 2.1.1.. 6+12

1...Bh5 2.gxh5 Qd6 + 3.Ke2 Re4#

1...Rh4 2.gxh4 Qd2 + 3.Kc4 Bd5#

Zilahi with elegant sacrifices leading to line opening for the WQ (PE)

3338

Menachem Witztum

Tel Aviv

H#2.5 2.1.1.. 11+8

1...Rxb3 2.Qxd4 Rxc3 3.Qe5 fxe5#

1...Sxc3 2.Qc6 Sb1 3.Qe4 fxe4#

Line opening for the BQ and exchange of function of the WR & WS (PE)

'dqd'dnd d'd'd'd' 'd'dbd'0 d'4'd'Gk 'd'd'dNd d'd'd'dr 'd'd'd'h g'd'd'!K

'd'd'd'd d'I'd'd' 'db4p0'd dpdkdpd' '0'd'g'd dPd'd'd' 'd')'d'H d'4'd'd'

'd'd'd'd d'd'dpd' K0'0')'d dPirdpd' '4nd'd'd dpd'0'd' 'GRd'dPd d'd'd'd'

'g'd'd'd d'd'd'0r 'd'd'dp! dn0'd'0' Rd'd'd'd d'dkdB)K 'd'd'0Ph d'd'4bd'

'd'd'd'I d'd'd'd' 'd'd'0'G dNd'dPd' q$')')'d )p0PiP$' 'dpdp0'd d'd'd'd'

10

3339

Aleksey Ivunin

Alexander Pankratiev

Russia

H#2.5 3.1.1.. 4+4

b)Bf1g1

a) 1...Ba6 2.Rxe4 Rxe4 + 3.Kd5 Bxb7#

1...Bc4 2.Rd1 Bd5 3.Rd4 Rf5#

1...Rf8 2.Ke6 Re8 + 3.Kd7 Bb5#

b) 1...Rf2 2.Kd4 Ka5 3.Kc5 Rc2#

1...Rf7 2.d5 Re7 + 3.Kd6 Bc5#

1...Rf6 2.d5 Kc5 3.dxe4 Bd4#

Many arrangements of mating positions, but I don’t see any unity (PE)

3340

Mirko Degenkolbe

Udo Degener

Germany

H#2.5 2.1.1.. 4+8

1...0-0-0 2.0-0 Rdg1 + 3.Kh7 Rxh3#

1...0-0 2.Bb1 Raxb1 3.0-0-0 Rc1 #

Four castlings, two in each solution, with similar mating positions in spite of the

differences of the castling side (PE)

3341

Viktor Yuzyuk

Ukraine

H#2.5 4.1.1.. 4+9

1...Ra3 + 2.Kd4 Ka7 3.c5 Qf4#

1...Ra5 2.Qg5 Qh8 3.Kf4 Qe5#

1...Bf3 2.Qh4 Qxb4 3.Qf2 Ra3#

1...Qxd8 2.Kf2 Qh4 + 3.Kg1 Ra1#

Exchange of functions (bPc7/bQd8, Line opening + Self-block / Passive)

Exchange of functions (wRa7/wQb8, Guard / Mate) ª 4 (author)

'd'd'd'd dpd'd'd' 'I'0'd'd d'd'i'd' 'd'dP$'d d'd'd'd' 'd'd'd'd d'd'4Bd'

rd'dkd'4 d'dpd'd' 'd'd'd'd d'd'd'd' 'd'dbd'd d'd'0'dp pd'dPd'd $'d'I'dR

K!'1'd'd $B0pd'd' 'd'dpd'd d'd'dpd' '0'd'dpd d'd'i'd' 'd'd'd'd d'd'd'dn

11

3342

Daniel Papack

Germany

H#3 2.1.1.1 5+11

1.Sxc3 Rxf2 + 2.Kxd1 Kb2 3.Se2 Rf1#

1.Sxf4 Rxd2 + 2.Kxf1 Kb1 3.Se2 Rd1#

A major point are the tries:

1.Sxc3 Rxd2 +? 2.Kxf1 Kb1 3.Se2 Rd1and no mate as the line of BBa5 is open.

Similarly:

1.Sxf4 Rxf2 + 2.Kxd1 Kb2 3.Se2 Rf1 and no mate as the line of te BQ is open.

(PE)

3343

Aleksey Ivunin

Alexander Pankratiev

Russia

H#3* 7+7

1...Sxe5 2.Kxe5 Rxf6 3.Kxf6 Se8#

1.Bxg7 Sxe5 2.Kxe5 Rh7 3.Kf6 Bxg7#

Capture of two white pieces in both set and solution, but the repeating moves

detract from the quality (PE)

3344

Mykola Kolesnik

Roman Zalokotsky

Ukraine

H#3 2.1.1.1 4+10

1.Bf5 Rxe8 2.Bg6 Sh3 + 3.Kf5 Bc8#

1.Re5 Bxc8 2.Rd5 Se2 + 3.Ke5 Re8#

Exchange of white’s 1st & 3rd moves with black performing similar strategy (PE)

3345

Misha Shapiro

Maale Edomim

H#3 b)Rf4g4 8+15

a) 1.Rxg3 Rf3 2.Rh3 Rxc3 3.Rd3 Rc4#

b) 1.Bxg3 Bf4 2.Bh2 Bxc7 3.Be5 Bb6#

Nice Turtons with tempo moves by black (PE)

'd'4'd'd d'd'dqd' 'd'd'dnd g'd'd'd' nd'd')'0 d')pd'd' 'd'0k0rd I'dRdRd'

'd'd'd'G d'd'0nH' 'd'dpgN$ d'd'0'0P 'd'iPd'd d'd'd'd' 'd'd'd'd I'd'd'd'

'db$rd'd dBd'd'g' 'dp0'0'd d'0'd'd' 'd'd'i'd d'dpd'd' 'd'd'd'd I'd'h'H'

'd'd'd'd d'0'd'd' 'dPd'0'd d'd'gP0' '0'iP$pd dP0rd')' 'dpG'dpd h'I'dn4b

12

33446

Misha Shapiro

Maale Edomim

H#3 2.1.1.1 6+4

1.Qg4! (Tempo, Qxg3?) Rf3 (a) 2.Qg3! (Tempo, Qh4?) e4 (b) 3.Qh4! Rf5#

1.Kf5! (Tempo, Kf4?) e4 +(b) 2.Kf4! (Kg5?) Rf3 +(a) 3.Kg5! Rf5#

Black can perform two triangular maneuvers reaching the same mating position.

BK in check is the way the composer managed to realize this (PE)

3347

Aleksey Ivunin

Alexander Pankratiev

Russia

H#3.5 3.1.1.. 3+9

1...Rxc3 2.Kf5 Rxc4 3.Kg4 Rxe4 + 4.Kh3 Rh4#

1...Bg3 2.Kd7 Rxh1 3.Kc8 Rb1 4.Qd7 Rb8#

1...Bf2 2.Kd5 Bg1 3.Qe6 Rxc3 4.Sd6 Rc5#

Echo mating positions with the white half-pin existing not for strategy but mostly

for forcing move order (PE)

3348

Mykola Kolesnik

Ukraine

H#3.5 b)Bh4=R 3+13

a) 1...Bf2 2.Qd3 Ba7 3.Rb6 Se3 + 4.Kd4 Bxb6#

b) 1...Rh6 2.Rf7 Ra6 3.Qb6 Sf6 + 4.Ke6 Rxb6#

Critical play enabling the Maslar theme and nice model mates (PE)

3349

Aleksey Ivunin

Alexander Pankratiev

Russia

H#5 2.1.1.1 2+8

1.Bc7 Bxb6 2.0-0-0 Kc4 3.Ba8 Kb5 4.Kb8 Ka6 5.Rc8 Ba7#

1.c4 + Kxd4 2.Rc8 Ke5 3.Bc7 + Ke6 4.Kd8 Bb4 5.Be8 Be7#

Echo mates but apart for many expected self-blocks there are no unifying

strategic elements (PE)

3350

Anatoly Styopochkin

Russia

H#5.5 2+7

1...Bxa6 2.Bb6 Be2 3.0-0-0 Kd2 4.Kb7 Kd3 5.Ka6 Kc4 6.Qb7 Kb4#

Annihilation Castling (black, long) Delayed Umnov Indian (white) Kniest theme

Switchback (B, with captures, 1) Battery mate Royal battery mate (author)

The parallel moves of the bishops and switchback make an interesting solution

(PE)

'd'd'd'd d'd'0'd' 'd'd'd'0 d'd')'iP 'd'd'd'1 d'd'd'$P 'd'dPd'd d'I'd'd'

'd'd'd'd d'd'd'd' 'd'1kd'd d'd'd'd' 'dndpd'd d'0pd'd' 'd'4'dpd I'$'G'dr

'd'd'd'd d'd'0'gp 'd'd'4'd d'dk0r0' 'd'dp0NG d'0'1bd' 'd'd'd'I d'd'd'd'

rd'gkd'd d'd'd'd' 'hbd'd'd G'0'd'dn 'd'0'd'd d'dKd'd' 'd'd'd'd d'd'd'd'

rd'dkd'd 0'd'd'd' pd'0'd'd d'd'd'd' 'd'd'd'd d'd'g'd' 'd'dBdqd d'd'I'd'

13

3351

Krzysztof Drążkowski

Poland

H#9 2+5

1.h5 Bd4 2.h4 Bxf2 3.h3 Bg1 4.h2 Kg2 5.h1=B + Kf2 6.Ka7 Ke3 7.Ra8 Kd4

8.Rb8 Kc5 9.Bb7 Kb5#

Only a bishop can do the job and block on b7. Nice precision of moves up to the

bishop promotion and white battery build-up, followed by three pleasing self-

blocks that must be carried in exact order (PE)

3352

Anatoly Styopochkin

Russia

S#3 12+10

1.Qe2 ! [2.Sxb3 + (A) cxb3 3.Qb5 + Bxb5#]

1...Sd3 2.Qf2 + Sxf2 3.Sxb3 + (A) cxb3#

1...Bxc1 2.Qxc4 + (B) Kxc4 3.axb3 + Kc5#

1...Rxb7 2.Rxc6 + Kxc6 3.Qxc4 + (B) Bxc4#

SOTF (self mate of the future)

The idea is that the threat and 1st variation form one pair and the other two

variations a 2nd pair. I find this of only limited success (PE)

3353

Hartmut Laue

Germany

S#3* 11+9

1...f6 a 2.Qc4 + A bxc4 3.Bxe6 c3#

1...fxg6 b 2.Bxe6 B zz g5 3.Qe3 + fxe3#

1.Bg5 ! zz.

1...f6 a 2.Bxe6 B zz. fxg5 3.Qe3 + fxe3#

1...fxg6 b 2.Qc4 +A bxc4 3.Bxe6 c3#

Reciprocal changes in mutate form. When black has no move available white

provides it with 2.Qc4 and when black has a move white prepares the sacrifice on

e3 with 2.Bxe6 (PE)

3354

Stanislav Vokal

Slovakia

S#5 8+5

1.Ba6! Sf1(Sf3) 2.Qd8+ Bc7 3.Qd2+ Sxd2 4.Re5+ Bxe5 5.b4+! axb3 e.p.x !

1.Bd3?, 1.Be2?, 1.Bf1?

Good tries and nice finale (PE)

3355

Udo Degener

Germany

S#8* 6+3

1...hxg3#

1.Qb1! hxg3+ 2.Kg1 Kh4/h5 3.Qh7+ Kg4 4.Rh2! gxh2+ 5.Kh1 Kg3 6.Qh5! g4

7.Qh8! gxh3 8.Qd4! hxg2#

A nice mutate with excellent maneuver of the WQ (PE)

kd'd'd'4 d'd'd'd' 'd'd'd'0 drd'd'd' 'd'd'd'd d'd'd'd' 'd'd'1'd G'd'd'dK

'4'd'd'd )B$'d'd' KdpdPd'd G'i'dNd' 'dpdPd'd dp0'd'd' PgPd'dpd d'HQhbd'

'd'd'dBd d'd'dp)' 'd'dpdPG dp!'0pd' ')'dk0'd d'dpdNd' 'd'IndPd d'dR$'d'

'd'd'd'd d'd'd'd' 'd'd'dRd i'd'g'!' pdBd'dpd )'d'd')' K)'d'd'h d'd'$'d'

'd'd'd'd d'd'd'd' 'd'd'd'd d'd'd'0' 'd'd'dk0 d'd'd'HN 'd'd'dPI d'd'd'!R

14

3356

Menachem Witztum

Tel Aviv

HS#2 b)pf5e5 8+13

a) 1.Qxd4 Rhxh6 2.Qe3 + Se4#

b) 1.Qxg5 Bxa6 2.Qg4 + Sf5#

Interesting battery mates with line opening, capture unguards and Zilahi (PE)

3357

Menachem Witztum

Ricardo Vieira

Tel Aviv/Brazil

HS#2.5 6+11

b)qa8d8

a) 1...Qxe4 2.Sxf4 Rxf4 3.Sxe5 + Qxe5#

b) 1...Qxd3 2.Rxe2 Bxe2 3.Se3 + Qxe3#

A Nt (original?) idea of making a “battery” from like moving pieces. In one

solution the BR is allowed to be placed behind the BQ and in the other the BB is

laced behind it. In both cases, the BQ can now be forced to move away from the

line for the mate (PE)

3358

Hubert Gockel

Germany

#2vvv Breton 10+10

1.b3 ? [2.Re5 A#]

1...Sxf7[-bBa7] 2.Sb6# but 1...fxe3[-bBa7] a !

1.b4 ? [2.Rd4 B #] but 1...Bxc5[-bSd8] b !

1.Qxd7[-wPb2] ? [2.Qxd6[-wRg4] C #]

1...fxe3[-bBa7] a 2.Re5 A # 1...Bxc5[-bSd8] b 2.Rd4 B # 1...Bb8 2.Sb6#

1...Sb7 2.Bxe6[-wPg5] # 1...Sxf7[-bBa7] 2.Sb6# but 1...Rxd7[-bQg3] !

1.Qxd7[-wPg5] ! [2.Rg5#]

1...fxe3[-bBa7] a 2.Re5 A #

1...Bxc5[-bSd8] 2.Rd4 B #

1...Qh3 2.Qxd6[-wRg4] # 1...Sxf7[-bBa7] 2.Sb6#

In Breton chess when a piece is capture the same kind of piece from the capturing

side is removed from the board.

We have almost a regular Dombrovskis, but some specific Breton play that needs

explanation: after 1.Qxd7 white can choose if WPb2 or WPg5 will be removed.

Removing WPb2 allows the seemingly crude refutation 1...Rxd7 which removes

the BQ but no mate can be given. Why 1.Qxd7[-wPg5] is different? First,

because now we cannot threat 2.Qxd6 as removing WRg4 is impossible –

exposing the WK to check, and removal of WRe4 allows BK escape. Now, in the

Dombrovskis defense 1...Bxc5[-bSd8] the mate 2.Rd4 B # is possible since te

capture 2...Bxd4 is illegal as BRd6 cannot be removed. Interesting use of this

fairy condition (PE)

3359

Semion Shifrin

Nesher

#2v 8+5

Leo Q Vao B

1.Qee7 ? [2.Qfxa7 A #]

1...c3 a 2.Qc4 B #

1...axb6 b 2.Qea7 C #

but 1...a5 !

1.Kd5 ! [2.Qxc4 B #]

1...c3 a 2.Qexa7 C #

1...axb6 b 2.Qa7 A #

This light position hides a complex mechanism showing the Shedey cycle. The

main idea follows the defenses by the black pawns that are captured by leaping

Leo moves: a leaping Leo cannot give mate on the evacuated square. There are

additional line effects caused by the try and key moves. Excellent! (PE)

'd'dNdBd d'dp0PdP Rg'dkdr$ d'dpdph' 'd'h'd'd dpd'dpdr 'd'!bd'd d'd'd'I'

qd'd'd'd d'd'd'd' 'd'd'4Bd g'I'0'0' 'dNdRhk0 d'dNdb0' 'd'dpdPd d'd'd'd'

'd'hQd'd g'dpdBI' 'dp4pd'd d'Hkd')' 'dNdR0Rd d'd'Gp1' ')'d'd'd d'd'd'd'

'!'d'dBd 0'd'dQd' pG'd'd'd d'd'd'd' kGpI'd'd d'dp!'d' 'd')'d'd d'd'd'd'

15

3360

Semion Shifrin

Nesher

H#2 4.1.1.1 8+13

Take&Make Pao r Vao Bb

1.Qf3 Qxb1-e4 + 2.Kxe4-d5 gxf3-e4#

1.VAc5 Qxf2-f4 + 2.Kxf4-c7 dxc5-b6#

1.VAe5 Qxf2-f3 + 2.Kxf3-a3 dxe5-b2#

1.PAc5 Qxe6-f4 + 2.Kxf4-h6 dxc5-g5#

In each of the four solutions the BK capture the WQ to move into the square on

which it is mated and a WP captures to move to the mating square. In all cases the

BK cannot capture the WP as tis is illegal (PE)

3361

Udo Degener

Germany

H#2 ParrainCirce 4+10

2.1.1.1 b)kh6a8

a) 1.Sxb3 Sc3-e4[+wSd4] 2.Sc3 Sf5#

1.Ra8 Sd4 2.Sxc3 Sf5 [+wSe4] #

b) 1.Sxc3 Sb3-a5[+wSb5] 2.Sb3 Sc7#

1.Rf6 Sb5 2.Sxb3 Sc7 [+wSc5] #

In this unpin-based problem I especially like the self-unpin by the white knights

using the ParrainCirce effect (PE)

3362

Udo Degener

Germany

H#6 b)kd6c7 2+2

KoeKo

a) 1.c1=S Kd2 2.Sd3 Ke3 3.Se5 Kf4 4.Sf7 e8=S 5.Sg5 Kf5 6.Se4 Ke6#

b) 1.c1=Q Kc2 2.Qd2 Kc3 3.Qd7 e8=Q 4.Qc6 Qc8+ 5.Kb6 Qa6+ 6.Ka5 Kb4#

In a) the interesting point is that at the 6th move the BS as a choice of three moves

but only one works: 6.Sf7 will make 6...Ke6 illegal as the BK can capture it and

will be in contact with the BS. 6.Se6 will follow with 6...Ke6 7.Kd5,d4 The

interesting point is that the WK cannot capture the BK but any move by the BK

will make it capturable. In b) again with a matching pair of promotions the end is

also by a WK mating move but the story is very different (PE)

3363

Sergej Smotrov

Kazakhstan

S#13 5+3

Grasshopper <>

Main Plan: Early 1.Qc3+? 1.Qa6 Kb4 2.Sb3+ Ka3 3.Qxa8 Ka2 4.Sd5 Ka3

5.Sc1 Ka4 6.Qa6 Ka3 7.Gd1 Ka4 8.Kb1 Ka3 9.Ka1 Ka4 10.Gb1 Ka3 11.Sb6

Kb4 12.Qc4+ Ka3 Main Plan, 13.Qc3+ Ba5xc3# Partial computer test 1. Returns

of white figures 2. Logic maneuver Kc2 => a1 3. Aristocrat (author)

'gr4BGKd d'dp0pd' Pd'gnd'd d'd'dQd' '0p)'d'd d'd'i'd' Pd'd'1Pd dbd'd'd'

'4pd'd'g 0'd'd'dp 'd'd'dri d'dPd'dp 'd'd'd'd dNH'd'd' nI'h'd'd d'd'd'd'

'd'd'd'd d'd')'d' 'd'i'd'd d'd'd'd' 'd'd'd'd d'd'd'd' 'dpd'd'd d'dKd'd'

>d'd'd'd d'd'd'd' 'H'd'd'd g'd'd'd' 'dQd'd'd i'd'd'd' 'dKd'd'd d<H'd'd'

16

3364

Sergej Smotrov

Kazakhstan

HS#3.5 2.1.1.. 3+3

ProteanChess

1...Ke5 2.Rf4 gxf4=R 3.Rg6 R8f5 4.Re6 + Kxe6=rR #

1...g4 2.Kc4 gxf3=R 3.Rg5 R8f4 4.Re5 + Kxe5=rR #

In Protean Chess the capturing piece transforms into the same king of piece as the

captured one. Chameleon-echo mates in which the two white rook “become” two

black rooks (PE)

3365

Alberto Armeni

Italy

H=7 1+11

1.e1=S Kg7 2.Qa2 Kxf6 3.Rb2 Kxg5 4.Bb1 Kxh5

5.Sc2 Kg4 6.Sa1 Kf3 7.c2 Ke2 =

To get into a stalemate position the WK captures three BP’s while black locks his

pieces and waits for the WK to block d1 & d2 (PE)

3366

Alberto Armeni

Italy

#2 14+7

Neutral pieces QN

1.Rh5 ! [2.Rxd5#]

1...Nb4,Nf4 2.e3# 1...Nc3,Nf6 2.Sf5#

1...Ne3,Nb6 2.Sxb5# 1...Nc7 2.Bc5#

1...Rxc4 2.Sxb5# 1...Rc5 2.Be5# 1...Qh8 2.Rg4#

1...Bxc4 2.Sc2# 1...bxc4 2.Sb5#

The main idea is the almost complete neutral knight wheel, though its seven

possible moves lead only to 4 different mates (PE)

3367

Anatoly Styopochkin

Russia

HS#3 2.1.1.. 11+7

1.Ra7 Ra6 2.Rb8 Ra4 3.Ra5 Rxa5#

1.Rb8 Rc8 2.Ra7 Rh8 3.Rg8 Rxg8#

The same strategy, rather symmetric, with black zugzwang ending (PE)

'd'd'4'd d'd'd'd' 'd'd'd'd d'I'd'0' 'd'dkd'd d'd'dRd' 'd'd'd'd d'd'd'$'

'd'd'dKd d'd'd'd' 'd'd'0bd d'd'd'0p 'd'd'd'd 0p0'd'd' 'd'dp1rd d'i'd'd'

'drd'!'d d'd'0P$R 'd'GPd'd dpdNd'd' 'dPi'd'I Hbd'dpH' ')'dP)'h d'd'd'd'

Kd'd'dRd dNd'd'dp 'drd'd') $'d'dPd' '0')'d'd 0Pd'd'Hp Pd'd'!bi d'd'd'd'

17

3368

Anatoly Styopochkin

Russia

HS#2 b) -bb8 6+6

a) 1.axb8=R a1=R 2.Ra2 + Rxa2 #

b) 1.b8=B f1=B 2.Qg2 + Bxg2#

Matching minor promotions (PE)

3369

Ľuboš Kekely

Slovakia

Ser-H=16 7+3

1.Kg3 2.Kf2 3.Ke2 4.Kxd2 5.Ke3 6.Kf4 7.Kxg5 8.Kh6 9.g5

10.g4 11.g3 12.g2 13.g1=S 14.Sf3 15.Sxe5 16.Sc6 Kxc6 =

Only a knight, promoted at g1, can remove WPe5 and sacrifice on the right square

to be eliminated & block BPc7 (PE)

3370

Vito Rallo

Italy

H#3 2.1.1.. 4+6

ParrainCirce

1.Bxf3 Be5-c7[+wSd5] 2.Kxd5 cxd3 [+wSe6] 3.Sc6[+bSc5] Sf4#

1.Kd5 cxd3 2.Bxf3[+bSe4] Bxd4 [+wSe2] 3.Sd6[+bSc6] Sf4#

Good ParrainCirce play bringing all knight into place (PE)

3371

Eric Huber

Romania

H#2.5 b)kh3f3 2+3

Total Invisible 4

a) 1...Qf1 2.fxg3 Qf4 3.gxh2 + TI~*h2#

b) 1...Qc6 2.fxe3 TI~*g4 3.exd2 Qe4#

Comments and explanations (by author):

a) 2 white invisibles are captured on g3-h2

1 invisible is on g2. If it is white, it is still pinned after g3xh2

1 white invisible captures h2 -> this invisible could only be wQh1/wRh1 (the

existence of Sf1/Sf3 is excluded by move wQf1-f3)

b) 2 white invisibles are captured on e3-d2

1 black invisible e4 is captured by Qc6xe4

1 invisible g2/g3 captures bRg4-> this invisible must be a wRg2

Kg'd'd'd )Pd'd'd' 'd'd'd'd d'd'd'd' '0'dQd'd $'d'0'd' pd'dk0'd dRd'd'd'

'd'dNd'd d'0'd'0' 'd'd'd'd $'I')')' 'd'dBdkd d'd'd'd' 'd'G'd'd d'd'd'd'

'd'd'd'd d'd'd'd' 'd'dkdpd d'd'G'd' 'd'h'd'd d'0nINd' 'dPdbd'd d'd'd'd'

'd'd'd'd d'd'd'd' 'd'd'd'd d'd'd'd' 'd'd'0rd d'd'd'dk 'd'd'd'd d'!'d'I'

18

3372

Cornel Pacurar

Adrian Storisteanu Canada

H#2 b)pa7h3 3+4

PointReflection

a) 1.a6 Rd6 + 2.Kh3 Be3#

b) 1.h2 Bd6 + 2.Ka7 Re3#

PointReflection: When two pieces of any color stand on the squares which are

symmetric with respect to the central point of the chessboard, they exchange their

roles (i.e. powers of movement). For example, white bishop b6 and white rook g3

are on the symmetry axis so the bishop will move like rook and the rook like

bishop.

The problem shows nice orthogonal-diagonal correspondence with the black king

moving like a pawn at the mating position (PE)

3373

Michael Grushko

Kiriat Bialik

Ser-H#14 0+3+2

ABC PWC Take&Make

1.<xe3-e4[+Pc3] 2.Pc2 3.Pc2- c1=R 4.Re1 5.Pd2 6.Pxe1-h1=Q[+Rd2]

7.Rg2 8.<g4 9.f2 10.f1=S 11.Sh2 12.Kf3 13.Kxg2-e2[+Rf3] 14.Kxf3-

h3[+Re2] Rxh2-f3[+Ne2] #

Nice PWC & T&M play, enabled to be dual free by the ABC. The final position

seems to arise out of nowhere… (PE)

3374

Michael Grushko

Kiriat Bialik

H#9.5 Functionary 2+3

b)kf5d5 H#8.5

Grasshopper >

a) 1...Sf2 2.Re3 Be4 + 3.Kg4 Bd5 4.Kf3 Sg4 5.<f2 Se5 +

6.Kg2 Sd3 7.<h2 Se1 + 8.Kh1 Sc2 9.Re2 Sd4 10.Rd2 Se2 #

b) 1...Se3 + 2.Kc6 Sd5 3.Re3 Be4 4.<b6 Bf5 5.Re6 Se7 +

6.Kb7 Sc8 7.<b8 Sd6 + 8.Ka8 Sb5 9.Re5 Be4#

Interesting maneuvers using the FunctionaryChess condition in which pieces can

move only when attacked (observed) by an enemy piece (PE)

3375

Jaroslav Štúň

Slovakia

HS#3 5+0

b)Qe5h5 c)Nf6c4

Republican ParrainCirce

Neutral Eagle N Neutral Grasshopper Q

a) 1.Ng5xe5 Nd5-e4[+Qf4] 2.Ne5xf6 Qf4xf6[+Nf8] 3.Nf8-g6

[+Qg4][+bKh5] + Nf5-f3[+wKf5] #

1.. Nf5-e6[+Qd6] 2.Ne5xf6 Qd6xf6[+Nh6] 3.Nh6-f7[+Nd7] [+bKe8] +

Ne6-c6[+wKe6] #

b) 1.Qh5xf5 Nd5-f4[+Nh4] 2.Nh4xf4 Ng5xf4[+Ne3] 3.Nf6-g5[+Qg3]

[+bKh4] + Nf4-f2[+wKf4] #

1... Qg5-g7[+Nf7] 2.Nf6-e7 Qf7-g5 3.Nd5-f6[+bKh6] + Nf6-f8[+wKf6] #

c) 1.Qe5-c5 Nd5-d3 2.Nf5-c6 Ng5xc4 3.Nc6-b5[+Nb3][+bKa4] + Nc4-

c2[+wKc4] #

1.Nf5-e4 Ng5xe4 2.Qe5-e3 [+Ne2] Ne4xc4 3.Nd5-d3[+Nc2] [+bKd1] +

Nd3b3[+wKd3] #

In Republican chess a mate is given when either side, having played a move, can

place the opposing king on a square on which it is legally mated. The Eagle is

much like a grasshopper but deflects 900 to either side after jumping over the

hurdle.

The same intricate mating position is repeated six times, four with the kings on

white squares and twice on black squares (echo & chameleon-echo) (PE)

'd'd'd'd 0'd'd'd' 'G'd'd'd d'd'd'd' Kd'd'd'd 0'dpi'$' 'd'd'd'd d'd'd'd'

'd'd'd'd d'd'd'd' 'd'd'd'd d'd'd'd' 'd'd'i'd d'?P)pd' 'd'd'd'd d'd'd'd'

'd'd'd'd d'd'd'd' 'd'd'?'d d'd'dkd' 'd'drdNd d'd'dBd' 'd'd'd'd d'd'd'd'

'd'd'd'd d'd'd'd' 'd'd'H'd d'dN!NH' 'd'd'd'd d'd'd'd' 'd'd'd'd d'd'd'd'

19

3376

Jaroslav Štúň

Slovakia

HS#4 b)Sb2c4 4+1

c)Bf4d1 d) c+Sb2f2

KoeKo SiameseKings

a) 1.Kf5 Rb6 2.Sc4 Rb6-c6+ 3.Kd6 Rc7 4.Ke5 Rc7-c5‡

b) 1.Bd2 Rb4 2.Sb3 Rb5 3.Be3 Rb6 4.Kd4 Rb5-b4‡

c) 1.Ba4 Rb3 2.Bb5 Re3 3.Kd3 Re2 4.Kd3 c4 Re2-c2‡

d) 1.Kc4 Rb3 2.Bc2 Rb5 3.Kc3 Rf5 4.Kd3 Rf5-f3‡

In SiameseKings the inevitable capture of either king constitutes a

mate. For example, a move that gives a check to both kings, from

which both cannot escape, is a mate.

Here we have nice chameleon-eco mates (PE)

3377

Luboš Kekely

Slovakia

Ser-H=8 b)+ng8 3+3

ChecklessChess

a) 1.e6 2.exd5 3.d4 4.d3 5.d2 6.d1=B 7.Bh5 8.g4 Ke8 =

b) 1.e5 2.e4 3.e3 4.e2 5.e1=R 6.Rh1 7.Rh8 8.Kh7 Qxg5 =

Miniature. Double excelsior. Minor promotions (author)

Two nice stalemate positions based on the fact that black cannot give

a check (PE)

3378

Sébastien Luce

France

Sh#6 b)Pa1c2 4+1

SymmetryCirceHorizontal

Neutral pawns P

a) 1.Kxd1(Nd8) 2.Nc6 3.Kxc1(Qc8) 4.Kb1 5.Na7 6.Kxa1(Ra8)

Nxc8(Qc1)#

b) 1.cxd1=B(Bd8) 2.Bg5 3.Bxc1(Qc8) 4.Qxc1(Bc8) 5.Bcg4

6.Bh5 Bxh5 (Bh4)#

A feast of promotions: when a pawn is captured on the 1st row it is

reborn on the 8th row (according to this circe rule) and is

immediately promoted. So we have promotions on both the 1st and

8th row, in b) in a single move (!) (PE)

'd'd'd'd d'd'd'd' 'd'd'd'd drI'd'd' 'd'dKG'd d'd'd'd' 'H'd'd'd d'd'd'd'

'd'I'd'd d'd'0'd' 'd'd'dkd d'dPd'0' 'd'd'd'd d'd'd'!' 'd'd'd'd d'd'd'd'

'd'd'd'd d'd'd'd' 'd'd'd'd d'dKd'd' 'd'd'd'd d'd'd'd' 'd'd'd'd )')Pi'd'